Taking
a bearingAs an area of application, consider navigation. An important task of navigation is the determination of a ship's position. We will assume that all distances are not affected by Earth's curvature, which therefore can be neglected; the following considerations can then be based on plane triangles.
A ship's compass differs from a terrestrial one in that the magnet is fixed on a compass-card, so that its markings indicate directions. Its casing has a mark, which is lined up exactly with the ships longitudinal axis. Using it, you can read off the direction of the ship's course. Instead of a magnetic compass, which is affected by the ship's iron, a gyro-compass is used these days or even satellite navigation.
For the determination of directions, the compass has an appliance, which is connected to the casing and can be turned about a vertical axis. There are holes at C and D, through which you focus an object and can read the direction on the compass. This gadget may even be a telescope. This act of determination of a direction is called "taking a bearing".
The direction AB is
called N 30º E, that is, you rotate AN about A from the North
30º to the East, in order to obtain this direction. W 40º S
yields the angle WAC.
Task of coastal navigation: Take as quickly as possible two bearings of two point A and B on land, the distances and directions of which you have obtained from a chart, compute the distances of the ship S from A and B!
According to the chart, AB = 10 km and g = 30º, while the points A and B are in the directions N 20º W (a) and N 50º W (b).
Solution: N1S1
represents North-South. You see that 
BSA = b - a, S2AS
= a (as alternate angle on parallels), BAS2
= g, BAS = g + a,
ABS = 180º - (b + g).
Since you know in the triangle ABS one side and the angles, it can be computed by the II. basic task.
| a/sin (b - a) = AS/sin[180 - (g + b) |
| a/sin (b - a) = BS/sin (g + a) |
| AS = asin (g + b)/ sin (b - a) = = 10·sin 80º/ sin 30º = 19.80 |
| BS = asin (g + a)/ sin (b - a) = = 10·sin 50º/ sin 30º = 14.32 |
The distances of the ship S from A and B are 19.80 km and 14.32 km respectively.
Since there exist today also radio stations and satellites, from the positions of which you can take bearings, the same calculation still applies. In practice, it is often sufficient to enter the bearings on the chart and to find the ship's position graphically.
The log is a metal cylinder with fins, which is linked by a long cable to a rotation counter at the stern of a ship. When the ship moves forward, the fins cause the cylinder to turn. Its rotation is transmitted through the cable to the recorder. Similarly as with road vehicles, you can now tell how far the ship has moved. Also, the number of rotations of a ship's propeller allow you to estimate the distance covered.
However, if you can take bearings from a wireless station or a point on land like a tower, these angles determine only similar triangles. You always require one length for the calculation of distances; such a length can come from the log.
Task: The direction of travel g of a ship is given. At a definite time, a first bearing a of the lighthouse A is taken and after a certain distance, measured by the log, a second bearing b. How far away from the lighthouse is the ship at the point S?
| a = N 30º W | b = N 40º W | S1S2 = 2 km | g = N 30º E |
Solution: You
know in the triangle AS1S2 :
a. AS2S1
= 180º - (b + g).
b. AS1S2 = a + g, c. S1AS2
= b - a
a/sin (b - a) =
AS2/sin (a + g)
AS2 = asin(a + g)/sin(b - a) =
2sin 60º/sin10º = 9.97.
The ship at S2 is 9.97 km from A.
In nautical language, distances are often given in nautical miles (1852 m).